Quantization and experimental realization of the Colonel Blotto game


We present a quantum mechanical version of the Colonel Blotto game, where two players, Blotto and Enemy, collocate their soldiers (resources) sequentially in a finite number of territories. We analyse the representative classical cases of this game as well as the trivial case—which on its turn has no interest at all in the point of view of classical game theory—where, surprisingly, a player that could control a single parameter can win the game even if he/she is greatly outnumbered by his/her opponent. Besides the theoretical study we present an experimental realization of classical game by using linear optics circuits as well as a proposal of an experimental investigation of the quantized game. Finally, in order to check our quantization scheme we also present computer simulation results.

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The authors gratefully acknowledge the financial support of Brazilian’s agencies CAPES, FAPERJ, CNPq and INCT—Quantum Information. This study was financed in part by the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior—Brasil (CAPES)—Finance Code 001.

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Correspondence to A. G. M. Schmidt.


Appendix A: \({{\hat{J}}}\) Operator for \(n=2\)

In this “Appendix” we show the important special case for the quantum Colonel Blotto game for two territories, namely \(n=2\). In order to obtain the entanglement operator \({\hat{J}}\) we need, in the first place, to write the operator \({\hat{A}}\). Choosing the sign of the first element of the second matrix to be positive and the second one to be negative yields,

$$\begin{aligned} {\hat{A}} = \left( \begin{array}{cc} 0 &{}\quad 1 \\ -\,1 &{}\quad 0 \\ \end{array} \right) \;\;\; \otimes \;\;\; \left( \begin{array}{cc} i &{}\quad 0 \\ 0 &{}\quad -\, i \\ \end{array} \right) , \end{aligned}$$

so, using Eq. (17)

$$\begin{aligned} {\hat{J}}= & {} \cos (\gamma /2)\; \left( \begin{array}{cccc} 1 &{}\quad 0&{}\quad 0&{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0&{}\quad 0\\ 0 &{}\quad 0 &{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\\ \end{array}\right) + i \sin (\gamma /2)\; \left( \begin{array}{cccc} 0 &{}\quad 0&{}\quad i&{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0&{}\quad -\,i\\ -\,i &{}\quad 0 &{}\quad 0&{}\quad 0\\ 0&{}\quad i&{}\quad 0&{}\quad 0\\ \end{array}\right) \nonumber \\= & {} \left( \begin{array}{cccc} \cos (\gamma /2) &{}\quad 0&{}\quad -\,\sin (\gamma /2)&{}\quad 0 \\ 0 &{}\quad \cos (\gamma /2) &{}\quad 0&{}\quad \sin (\gamma /2)\\ \sin (\gamma /2) &{}\quad 0 &{}\quad \cos (\gamma /2)&{}\quad 0\\ 0&{}\quad -\,\sin (\gamma /2)&{}\quad 0&{}\quad \cos (\gamma /2)\\ \end{array}\right) . \end{aligned}$$

This is the same matrix as \({\hat{R}}(\gamma /2)\otimes {\hat{\varPi }}_1+{\hat{R}}(-\gamma /2)\otimes {\hat{\varPi }}_2\), which corresponds to rotations in different battlefields, namely

$$\begin{aligned} \left( \begin{array}{cc} \cos (\gamma /2) &{}\quad -\,\sin (\gamma /2) \\ \sin (\gamma /2) &{}\quad \cos (\gamma /2) \\ \end{array} \right) \otimes \left( \begin{array}{cc} 1 &{}\quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) + \left( \begin{array}{cc} \cos (\gamma /2) &{}\quad \sin (\gamma /2) \\ -\,\sin (\gamma /2) &{}\quad \cos (\gamma /2) \\ \end{array} \right) \otimes \left( \begin{array}{cc} 0 &{}\quad 0 \\ 0 &{}\quad 1 \\ \end{array} \right) ,\nonumber \\ \end{aligned}$$

which is the same as Eq. (25). For the case maximally entanglement \(\gamma = \pi /2\) then,

$$\begin{aligned} {\hat{J}} = {\hat{R}}(\pi /4)\otimes {\hat{\varPi }}_1+{\hat{R}}(-\,\pi /4)\otimes {\hat{\varPi }}_2. \end{aligned}$$

Appendix B: Classical Nash equilibrium

In this “Appendix” we will illustrate our quantized version of Colonel Blotto game showing detailed calculations which lead to the results presented in Table 1. In this example Enemy chooses the strategy referring to Nash equilibrium and Blotto chooses an arbitrary strategy. We are considering the specific classical case with three battlefields, both players have six soldiers and they are indivisible. The strategy that achieves the classical Nash equilibrium is the mixed strategy composed by five pure strategies given by Table 1, each one with probability equal to 1 / 5.

Since all classical strategies commute with entanglement operator \({\hat{J}}\), the final state is,

$$\begin{aligned} \vert \psi _f \rangle = {\hat{U}}_B {\hat{U}}_E \vert \psi _i \rangle . \end{aligned}$$

Enemy acts first and he applies his strategy to the initial state,

$$\begin{aligned} {\hat{U}}_E \vert \psi _i \rangle = \sum _{j=1}^3 {\hat{R}}(\lambda _j^E)\otimes {\hat{\varPi }}_j \left( \frac{1}{\sqrt{2}} \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \frac{1}{\sqrt{3}}\sum _{k=1}^3 \vert T_k \rangle \right) , \end{aligned}$$

which yields

$$\begin{aligned} {\hat{U}}_E \vert \psi _i \rangle = \frac{1}{\sqrt{6}} \sum _{j=1}^3 \left( {\hat{R}}(\lambda _j^E) \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_j \rangle \right) , \end{aligned}$$

when the Enemy chooses the strategy \(s_1\) from Table 1—zero, four and two soldiers dispatched to battlefields one, two and three, respectively—the angles which will rotate the initial state are given by,

$$\begin{aligned} \lambda _1^E = 0 ; \qquad \lambda _2^E=-\frac{\pi }{4} \frac{4}{6}=-\frac{\pi }{6}; \qquad \lambda _3^E=-\frac{\pi }{4} \frac{2}{6}=-\frac{\pi }{12}, \end{aligned}$$

substituting the angles in Eq. (30) we obtain,

$$\begin{aligned} {\hat{U}}_E \vert \psi _i \rangle= & {} \frac{1}{\sqrt{6}} \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_1 \rangle +{\hat{R}}(-\frac{\pi }{6}) \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_2 \rangle )\nonumber \\&+\, {\hat{R}}\left( -\,\frac{\pi }{12}\right) \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_3 \rangle \end{aligned}$$

now Blotto applies an arbitrary strategy,

$$\begin{aligned} \displaystyle {\hat{U}}_B{\hat{U}}_E \vert \psi _i \rangle= & {} \frac{1}{\sqrt{6}} {\hat{R}}(\lambda _1^B) \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_1 \rangle \nonumber \\&+\,{\hat{R}}(\lambda _2^B-\frac{\pi }{6}) \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_2 \rangle )\nonumber \\&+\, {\hat{R}}(\lambda _3^B-\frac{\pi }{12}) \left( \vert 0\rangle +\vert 1 \rangle \right) \otimes \vert T_3 \rangle \end{aligned}$$

Now we apply the rotation matrix to the state \(\vert 0\rangle +\vert 1 \rangle \)

$$\begin{aligned} {\hat{R}}(\theta ) \left( \vert 0\rangle +\vert 1 \rangle \right) = \cos (\theta )\vert 0\rangle + \sin (\theta )\vert 1 \rangle \end{aligned}$$

Using the above equation and the pay-off definition we have,

$$\begin{aligned} \$^1_B= & {} \mathrm{sgn}\left( \frac{1}{6}\Vert \sin (\lambda _1^B) \Vert ^2 - \frac{1}{6} \right) + \mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _2^B-\frac{\pi }{6}\right) \right\| ^2 - \frac{1}{6} \right) \nonumber \\&+\, \mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _3^B-\frac{\pi }{12}\right) \right\| ^2 - \frac{1}{6} \right) . \end{aligned}$$

In the last equation we calculated Blotto’s pay-off when Enemy chooses the strategy \(s_1\). In order to investigate the Nash equilibrium we need to calculate the pay-off for all others strategies, namely (\(s_2\), \(s_3\), \(s_4\), \(s_5\)), and the procedure is analogue. Therefore, after we calculate all these pay-offs it is necessary to find the expected pay-off which is defined as the sum,

$$\begin{aligned} \$_B = \sum _{i=1}^5 p_i \$_B^i \end{aligned}$$

where \(\$_B^i\) is Blotto’s pay-off for each strategy \(s_i\) chosen by his enemy, and all of them have probability 1 / 5. The idea is to test if Blotto can win with an arbitrary strategy against the best classical strategy (which is the strategy related to Nash equilibrium). So we have for strategy \(s_2\),

$$\begin{aligned} \$ _B^2= & {} \mathrm{sgn}\left( \frac{1}{6}\left\| \sin \left( \lambda _1^B-\frac{\pi }{24}\right) \right\| ^2 - \frac{1}{6} \right) +\mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _2^B-\frac{\pi }{24}\right) \right\| ^2 - \frac{1}{6} \right) \nonumber \\&+\,\mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _3^B-\frac{\pi }{6}\right) \right\| ^2 - \frac{1}{6} \right) , \end{aligned}$$

strategy \(s_3\) gives a pay-off,

$$\begin{aligned} \$ _B^3= & {} \mathrm{sgn}\left( \frac{1}{6}\left\| \sin \left( \lambda _1^B-\frac{\pi }{12}\right) \right\| ^2 - \frac{1}{6} \right) +\mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _2^B-\frac{\pi }{8}\right) \right\| ^2 - \frac{1}{6} \right) \nonumber \\&+\,\mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _3^B-\frac{\pi }{24}\right) \right\| ^2 - \frac{1}{6} \right) , \end{aligned}$$

the fourth strategy \(s_4\),

$$\begin{aligned} \$ _B^4= & {} \mathrm{sgn}\left( \frac{1}{6}\left\| \sin \left( \lambda _1^B-\frac{\pi }{8}\right) \right\| ^2 - \frac{1}{6} \right) +\mathrm{sgn}\left( \frac{1}{6} \Vert \sin (\lambda _2^B) \Vert ^2 - \frac{1}{6} \right) \nonumber \\&+\,\mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _3^B-\frac{\pi }{8}\right) \right\| ^2 - \frac{1}{6} \right) , \end{aligned}$$

and for the last one, \(s_5\), Blotto obtains

$$\begin{aligned} \$ _B^5= & {} \mathrm{sgn}\left( \frac{1}{6}\left\| \sin \left( \lambda _1^B-\frac{\pi }{6}\right) \right\| ^2 - \frac{1}{6} \right) +\mathrm{sgn}\left( \frac{1}{6} \left\| \sin \left( \lambda _2^B-\frac{\pi }{12}\right) \right\| ^2 - \frac{1}{6} \right) \nonumber \\&+\,\mathrm{sgn}\left( \frac{1}{6} \Vert \sin (\lambda _3^B) \Vert ^2 - \frac{1}{6} \right) \end{aligned}$$

Finally the expected pay-off is,

$$\begin{aligned} \$ _B = \frac{1}{5}\sum _{i=1}^5\$_B^i, \end{aligned}$$

using this final equation we plot the graphic in Fig. 5.

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Maioli, A.C., Passos, M.H.M., Balthazar, W.F. et al. Quantization and experimental realization of the Colonel Blotto game. Quantum Inf Process 18, 10 (2019). https://doi.org/10.1007/s11128-018-2113-5

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  • Quantum games
  • Optical circuits
  • Transverse modes
  • Polarization modes